Assume that $f:\mathbb{R}→\mathbb{R}$ is a Lipschitz function. Prove that if $g_n→g$ uniformly, then $f\circ g_n→f\circ g$ uniformly.

Question

Assume that $f:\mathbb{R}→\mathbb{R}$ is a Lipschitz function (which means that there exists a number $K\ge 0$ such that $|f(x)−f(y)| \le K|x−y|$, for all $x, y\in \mathbb{R}$). Also, suppose that functions $g_n, g:\mathbb{R}^d→\mathbb{R}$ are all bounded. Prove that if $g_n→g$ uniformly, then $f\circ g_n→f\circ g$ uniformly.

So if $g_n \rightarrow g$, then for any $\epsilon >0$ there exists an $N>0$ such that $n \ge N$ implies $|g_n-g|<\epsilon$. Not sure if this will be needed for the proof. If $g_n \rightarrow g$ , I don't see why you cant just say $f(g(x))_n\rightarrow f(g(x))$, since it is given. What does applying $f$ have to do with changing the fact that $g_n\rightarrow g$?

Answer 1

Hint : For all $x$, $$|f \circ g_n(x) - f \circ g(x)| \leq K |g_n(x) - g(x)|$$

so $$||f \circ g_n - f \circ g ||_{\infty} \leq K ||g_n - g||_{\infty}$$